Need some help with math

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  • #1
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A sequence [tex]t_{1}[/tex],[tex]t_{2}[/tex],...[tex]t_{n},...[/tex] is defined as follows:

[tex]t_{1}=14[/tex]
[tex]t_{k}=24-5t_{k-1}[/tex] for each k <= 2

For every positive integer [tex]n[/tex], [tex]t_{n}[/tex] can be expressed as [tex]t_{n}=p*q^n+r[/tex] where p, q and r are constants. The value of p, q and r is?

I found out [tex]t2=-46[/tex] and [tex]t3=254[/tex] and wrote out

[tex]14=p*q+r\\[/tex]
[tex]-46=p*q^2+r\\[/tex]
[tex]254=p*q^3+r[/tex]

I don't know what to do from here... is what I did so far even correct? TiA.
 

Answers and Replies

  • #2
arildno
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First of all, the recurrence relation ought to be:
[tex]t_{k}=24-5t_{k-1},k>=2[/tex]
Now, for arbitrary k, you have been given:
[tex]t_{k}=pq^{k}+r,t_{k-1}=pq^{k-1}+r[/tex]
Put this into the recurrence relation, and rearrange a bit:
[tex](q+5)pq^{k-1}=24-6r[/tex]
This equality must hold for EVERY k>=2. What does that suggest to you?
 
  • #3
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I don't really get what's happening...
 
  • #4
HallsofIvy
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Well, the first thing that happened is that you had a "misprint". You had written "for each k<= 2" when you clearly meant "for each k>= 2". That was what arildno was talking about.

THEN his point was that, if you know tn= p qn+ r, you can put that into the recursion formula. Replacing n by n-1 gives tn-1= p qn-1+ r so the recursion formula says p qn+ r= 24-5(p qn-1+ r). That equation is true for all n and has 3 unknown numbers in it so you can evaluate it for 3 values of n (2, 3, 4 will do) to get 3 equations for those 3 unknowns.
That's exactly what you did so, yes, what you have done so far is correct.

Now, to solve those equations:

An obvious thing to do is subtract the first equation from the second so that you get
60= pq- pq2=pq(1-q).

Subtract the second equation from the third to get 300= pq3- pq2= pq2(1-q).

Now, from the first equation, p= 60/(q(1-q)) so the second equation is
300= 60 (q2(1-q))/(q(1-q)). That is, 300= 60 q or q= 5.
Now work back to solve for p and r.
 
  • #5
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I got it... thanks for your help!

...but now I have another question that I'm stuck on. I can't even begin to do anything because I have no idea how to find out the answer... so some hints or something would be appreciated.

A circle that has a radius of 3 is tangent to both the positive x-axis and the positive y-axis. Another circle with a radius of 1 is tangent to the bigger circle and to the positive x-axis. There is a line tangent to both circles. Where does this line cross the y-axis?

I had to change the question around a bit... there's a picture associated that makes it eaiser to imagine but its not that hard to draw out.... and the line isn't the x-axis!
 
  • #6
xanthym
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Refer to the following diagram:
http://img110.exs.cx/img110/7287/tangentcircles4lh.jpg
From the RED Right-Triangle construct, determine the coordinates of Points A, B, C, and D. The equation of Line AC through points A and C can then be found, and thus the equation of Line ED through Point E and Normal to Line AC at Point D can be determined. The Y-intercept of Line ED is the solution to your problem.
 
  • #7
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Thanks for the help... I have another (simple) question: How do I evaluate (3^2000+3^-2000)^2-(3^2000-3^-2000)^2
 
  • #8
Curious3141
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Better to put different questions into different threads.

For the last one, use [tex]a^2 - b^2 = (a+b)(a-b)[/tex] and see what you get.
 
  • #9
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Ill do that in the future... I just dont know how to simplify (do I even simplify?) something like 3^2000+3^2000...

this is the solution... but I don't understand where 2.3 came from and how it becomes 4.3, etc...

http://img77.exs.cx/img77/8510/dme9zz.gif
 
  • #10
xanthym
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preet said:
Ill do that in the future... I just dont know how to simplify (do I even simplify?) something like 3^2000+3^2000...

this is the solution... but I don't understand where 2.3 came from and how it becomes 4.3, etc...

http://img77.exs.cx/img77/8510/dme9zz.gif

{2.32000} should be interpreted {2*32000}
{4.30} should be interpreted {4*30}


~~
 
Last edited:
  • #11
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wicked... it makes sense now =)
 

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